A. Vladik and Courtesy

在常规比赛中，Vladik 和 Valera 分别赢得了 #cf_span[a] 和 #cf_span[b] 颗糖果。Vladik 向 Valera 赠送了 #cf_span[1] 颗糖果。之后，Valera 为了不让别人觉得他不够慷慨，回赠了 Vladik #cf_span[2] 颗糖果。接着，Vladik 为了同样的原因，在下一轮赠送了 #cf_span[3] 颗糖果给 Valera。 更形式化地说，两人轮流赠送对方比上一轮收到的多一颗的糖果。 这个过程持续到其中一人无法送出规定数量的糖果为止。两人从对方那里收到的糖果 *不被视为他们自己的*。你需要知道，谁是第一个无法送出正确数量糖果的人。 输入数据仅有一行，包含两个用空格分隔的整数 #cf_span[a], #cf_span[b] (#cf_span[1 ≤ a, b ≤ 109])，分别表示 Vladik 和 Valera 拥有的糖果数量。 如果 Vladik 是第一个无法送出正确数量糖果的人，请输出一行 "_Vladik_"，否则输出 "_Valera_"。 第一个测试用例的图示： 第二个测试用例的图示： ## Input 输入数据仅有一行，包含两个用空格分隔的整数 #cf_span[a], #cf_span[b] (#cf_span[1 ≤ a, b ≤ 109])，分别表示 Vladik 和 Valera 拥有的糖果数量。 ## Output 如果 Vladik 是第一个无法送出正确数量糖果的人，请输出一行 "_Vladik_"，否则输出 "_Valera_"。 [samples] ## Note 第一个测试用例的图示： 第二个测试用例的图示：

Let $ a $ be the initial number of candies Vladik has, and $ b $ be the initial number of candies Valera has. The exchange proceeds in turns, starting with Vladik: - Turn 1 (Vladik gives): Vladik gives 1 candy to Valera. - Turn 2 (Valera gives): Valera gives 2 candies to Vladik. - Turn 3 (Vladik gives): Vladik gives 3 candies to Valera. - Turn 4 (Valera gives): Valera gives 4 candies to Vladik. - ... - On turn $ n $, the current player gives $ n $ candies to the other. **Important**: Candies received from the other player are **not** considered part of their own stock for the purpose of giving. That is, a player can only give candies from their **initial** stock — they cannot give candies they received in prior turns. Thus, at any turn $ n $, the player whose turn it is must have at least $ n $ candies from their original pile to give. Turns alternate: - Odd turns ($ n = 1, 3, 5, \dots $): Vladik gives. - Even turns ($ n = 2, 4, 6, \dots $): Valera gives. Define: - Let $ k $ be the turn number (starting at 1). - On turn $ k $, the player must give $ k $ candies. - Vladik can give on odd turns: $ k = 2m - 1 $ for $ m = 1, 2, 3, \dots $ - Valera can give on even turns: $ k = 2m $ for $ m = 1, 2, 3, \dots $ Vladik can make all his turns up to the largest odd integer $ k $ such that $ k \leq a $. Valera can make all his turns up to the largest even integer $ k $ such that $ k \leq b $. We need to find the **first** turn $ k $ where the player cannot give $ k $ candies. That is, find the smallest $ k \geq 1 $ such that: - If $ k $ is odd: $ a < k $ → Vladik fails. - If $ k $ is even: $ b < k $ → Valera fails. We are to determine the smallest $ k $ violating the condition, and report the player who fails on that turn. Thus, we compute: Let $ k_V $ be the largest odd integer such that $ k_V \leq a $ → Vladik can complete all odd turns up to $ k_V $. He fails on the next odd turn: $ k = k_V + 2 $, if $ k_V + 2 > a $. Actually, simpler: Vladik fails on the **first odd** $ k $ such that $ k > a $. That is, the smallest odd integer greater than $ a $. Similarly, Valera fails on the smallest even integer greater than $ b $. Let: - $ k_V = \min \{ k \in \mathbb{N} \mid k \text{ odd}, k > a \} $ - $ k_{Va} = \min \{ k \in \mathbb{N} \mid k \text{ even}, k > b \} $ Then: - If $ k_V < k_{Va} $, then Vladik fails first. - Else, Valera fails first. Compute: - $ k_V = \begin{cases} a + 1 & \text{if } a \text{ is even} \\ a + 2 & \text{if } a \text{ is odd} \end{cases} $ - $ k_{Va} = \begin{cases} b + 1 & \text{if } b \text{ is odd} \\ b + 2 & \text{if } b \text{ is even} \end{cases} $ Alternatively, more cleanly: - $ k_V = a + 1 + (a \bmod 2) $ - $ k_{Va} = b + 1 + (1 - b \bmod 2) $ But even simpler: The smallest odd integer > $ a $ is: $ k_V = a + 1 $ if $ a $ even, else $ a + 2 $ → $ k_V = a + 1 + (a \bmod 2) $ Similarly, smallest even integer > $ b $: $ k_{Va} = b + 1 $ if $ b $ odd, else $ b + 2 $ → $ k_{Va} = b + 1 + (1 - b \bmod 2) = b + 2 - (b \bmod 2) $ Alternatively, use: - $ k_V = 2 \left\lceil \frac{a + 1}{2} \right\rceil - 1 $ - $ k_{Va} = 2 \left\lceil \frac{b + 1}{2} \right\rceil $ But computationally, we can just compute: Let: - $ next\_odd\_after\_a = a + 1 $ if $ a $ even, else $ a + 2 $ - $ next\_even\_after\_b = b + 1 $ if $ b $ odd, else $ b + 2 $ Then compare: - If $ next\_odd\_after\_a < next\_even\_after\_b $ → Vladik fails first → output "Vladik" - Else → Valera fails first → output "Valera" **Formal Statement:** Given integers $ a, b \geq 1 $, define: - $ k_V = \begin{cases} a + 1 & \text{if } a \equiv 0 \pmod{2} \\ a + 2 & \text{if } a \equiv 1 \pmod{2} \end{cases} $ - $ k_{Va} = \begin{cases} b + 2 & \text{if } b \equiv 0 \pmod{2} \\ b + 1 & \text{if } b \equiv 1 \pmod{2} \end{cases} $ Then: - If $ k_V < k_{Va} $, output "Vladik" - Else, output "Valera"